Stoichiometry: Chemical Calculations Chemistry 120 Chemistry Is Concerned with the Properties and the Interchange of Matter by Reaction

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Stoichiometry: Chemical Calculations Chemistry 120 Chemistry is concerned with the properties and the interchange of matter by reaction – structure and change.

In order to do this, we need to be able to talk about numbers of atoms

Stoichiometry: Chemical Calculations Chemistry 120

The mole and atomic mass The mole is defined as the number of elementary entities as are present in 12.00 g of 12C. Numerically, this is equal to Avogadro’s Number 6.022 x 1023 Therefore, in 12.00 g of 12C there are 6.022 x 1023 ‘elementary entities’, in this case atoms.

Stoichiometry: Chemical Calculations Chemistry 120

The mole and atomic mass Atomic masses, in atomic units, u, are defined relative to 12C. Therefore, The formula mass of an element or compound contains 1 mole, 6.022 x 1023, of particles

1 Stoichiometry: Chemical Calculations Chemistry 120

Examples How many particles are there in 5 g of Na?

Stoichiometry: Chemical Calculations Chemistry 120

Examples How many particles are there in 5 g of Na? The particles are atoms – how many atoms are there in 5 g of Na?

Stoichiometry: Chemical Calculations Chemistry 120

Examples How many particles are there in 5 g of Na? The particles are atoms – how many atoms are there in 5 g of Na? Atomic mass of Na = 22.9898 u

2 Stoichiometry: Chemical Calculations Chemistry 120

Examples How many particles are there in 5 g of Na? The particles are atoms – how many atoms are there in 5 g of Na Atomic mass of Na = 22.9898 u As 1 12 1 u = /12 x mass ( C) And 1 mole = 6.022 x 1023 particles = number of particles in 12 g 12C

Stoichiometry: Chemical Calculations Chemistry 120

Examples How many particles are there in 5 g of Na? The particles are atoms – how many atoms are there in 5 g of Na Atomic mass of Na = 22.9898 u Mass of 1 mole of Na = 22.9898 g

Stoichiometry: Chemical Calculations Chemistry 120

Examples How many particles are there in 5.0000 g of Na? 22.9898 g Na = 1 mole Na Then 1 g Na = 1 mol Na 22.9898 5 x 1 g Na = 5 x 1 mol Na 22.9898 5 g Na = 0.2175 mol Na 5 g Na = 0.2175 x (6.022 x 1023) particles Na

3 Stoichiometry: Chemical Calculations Chemistry 120

Examples How many particles are there in 5.0000 g of Na? 1.310 x 1023 atoms

Stoichiometry: Chemical Calculations Chemistry 120

Examples What is the mass of 0.23 mol of butane?

Stoichiometry: Chemical Calculations Chemistry 120

Examples What is the mass of 0.23 mol of butane?

Molecular formula of butane: C4H10

4 Stoichiometry: Chemical Calculations Chemistry 120

Examples What is the mass of 0.23 mol of butane?

Molecular formula of butane: C4H10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g

Stoichiometry: Chemical Calculations Chemistry 120

Examples What is the mass of 0.23 mol of butane?

Molecular formula of butane: C4H10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g

Molecular mass of C4H10 = (4x12.011)+(10x1.0079)u = 58.123 u

Stoichiometry: Chemical Calculations Chemistry 120

Examples What is the mass of 0.23 mol of butane?

Molecular formula of butane: C4H10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g

Molecular mass of C4H10 = (4x12.011)+(10x1.0079)u = 58.123 u Relative Molecular Mass of Butane = 58.123 g

5 Stoichiometry: Chemical Calculations Chemistry 120

Examples What is the mass of 0.23 mol of butane? Relative Molecular Mass of Butane = 58.123 g 1 mole of butane = 58.123 g 0.23 x 1 mole of butane = 0.23 x 58.123 g

Stoichiometry: Chemical Calculations Chemistry 120

Examples What is the mass of 0.23 mol of butane? Relative Molecular Mass of Butane = 58.123 g 1 mole of butane = 58.123 g 0.23 x 1 mole of butane = 0.23 x 58.123 g

0.23 mole of butane = 13.368 g

Stoichiometry: Chemical Calculations Chemistry 120

Chemical Equations These are formulæ which show the chemical change taking place in a reaction.

Sr(s) + Cl2(g) SrCl2(s)

6 Stoichiometry: Chemical Calculations Chemistry 120

Chemical Equations These are formulæ which show the chemical change taking place in a reaction.

Sr(s) + Cl2(g) SrCl2(s)

Physical state

Stoichiometry: Chemical Calculations Chemistry 120

Chemical Equations These are formulæ which show the chemical change taking place in a reaction.

Sr(s) + Cl2(g) SrCl2(s)

Reactants Physical state Product

Stoichiometry: Chemical Calculations Chemistry 120

Chemical Equations As matter cannot be created or destroyed in a chemical reaction, the total number of atoms on one side must be equal to the total number of atoms on the other.

7 Stoichiometry: Chemical Calculations Chemistry 120

Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water

Stoichiometry: Chemical Calculations Chemistry 120

Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water

Reactants: Cyclohexane, C6H12

Oxygen, O2

Stoichiometry: Chemical Calculations Chemistry 120

Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water

Reactants: Cyclohexane, C6H12

Oxygen, O2

Products: Carbon Dioxide, CO2

Water, H2O

8 Stoichiometry: Chemical Calculations Chemistry 120

Chemical Equations Example Initially, we can write the reaction as

C6H12 + O2 CO2 + H2O

Stoichiometry: Chemical Calculations Chemistry 120

Chemical Equations Example Initially, we can write the reaction as

C6H12 + O2 CO2 + H2O This is NOT a correct equation – there are unequal numbers of atoms on both sides

Stoichiometry: Chemical Calculations Chemistry 120

Chemical Equations Example Initially, we can write the reaction as

C6H12 + O2 CO2 + H2O This is NOT a correct equation – there are unequal numbers of atoms on both sides Reactants: 6 C, 12 H, 2 O Products: 1 C, 2 H, 3 O

9 Stoichiometry: Chemical Calculations Chemistry 120

Balancing the equation

C6H12 + O2 CO2 + H2O

Stoichiometry: Chemical Calculations Chemistry 120 Balancing the equation C6H12 + O2 CO2 + H2O 6 C, 12 H, 2 O 1 C, 2 H, 3 O 6 C on LHS means there must be 6 C on the RHS

C6H12 + O2 6CO2 + H2O 6 C, 12 H, 2 O 6 C, 2 H, 13 O 13 O on RHS means there must be 13 O on LHS 13 C6H12 + /2 O2 6CO2 + H2O 6 C, 12 H, 13 O 6 C, 2 H, 13 O

Stoichiometry: Chemical Calculations Chemistry 120 Balancing the equation 13 C6H12 + /2 O2 6CO2 + H2O 6 C, 12 H, 13 O 6 C, 2 H, 13 O 12 H on RHS means there must be 12 H on LHS

13 C6H12 + /2 O2 6CO2 + 6H2O 6 C, 12 H, 13 O 6 C, 12 H, 18 O 18 O on RHS means there must be 18 H on LHS

C6H12 +9O2 6CO2 +6H2O 6 C, 12 H, 18 O 6 C, 12 H, 18 O

10 Stoichiometry: Chemical Calculations Chemistry 120 The final balanced equation is

C6H12 +9O2 6CO2 +6H2O and the coefficients are known as the stoichiometric coefficients. These coefficients give the molar ratios for reactants and products This is a stoichiometric reaction – one where exactly the correct number of atoms is present in the reaction

Stoichiometry: Chemical Calculations Chemistry 120 If cyclohexane were burnt in an excess of oxygen, the quantity of oxygen used would be the same although O2 would be left over.

Stoichiometry: Chemical Calculations Chemistry 120 The final balanced equation is

and the coefficients are known as the stoichiometric coefficients. These coefficients give the molar ratios for reactants and products This is a stoichiometric reaction – one where exactly the correct number of atoms is present in the reaction

11 The Exam Chemistry 120

Solutions Chemistry 120 A solution is a homogenous mixture which is composed of two or more components the solvent - the majority component and one or more solutes - the minority components

Solutions Chemistry 120 Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent.

Some are solids where both the solvent and the solute are solids. Brass is an example

12 Solutions Chemistry 120 Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent.

Some are solids where both the solvent and the solute are solids. Brass is an example

Solutions Chemistry 120 Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent.

Some are solids where both the solvent and the solute are solids. Brass is an example

Here copper is the Zn solvent, zinc the solute. Cu

Solutions Chemistry 120 Gas-Solid solution: Hydrogen in palladium

Steel

13 Solutions Chemistry 120 Common laboratory solvents are usually organic liquids such as acetone, hexane, benzene or ether or water. Water is the most important solvent. The oceans cover ~ ¾ of the surface of the planet and every cell is mainly composed of water.

Solutions in water are termed aqueous solutions and species are written as E(aq).

Solutions Chemistry 120 Aqueous Solutions Water is one of the best solvents as it can dissolve many molecular and ionic substances. The properties of solutions which contain molecular and ionic solutes are very different and give insight into the nature of these substances and solutions.

Solutions Chemistry 120 Ionic Solutions

An ionic substance, such as NaClO4, contain ions – + - in this case Na and ClO4 . The solid is held together through electrostatic forces between the ions. In water, the solid dissolves and the particles move away from each other and diffuse through the solvent. This process is termed Ionic Dissociation

14 Solutions Chemistry 120 Ionic Solutions In an ionic solution, there are therefore charged particles – the ions – and as the compound is electrically neutral, then the solution is neutral. When a voltage is applied to the solution, the ions can move and a current flows through the solution. The ions are called charge carriers and whenever electricity is conducted, charge carriers are present.

Solutions Chemistry 120 Molecular Solutions A molecular solution does not conduct electricity as there are no charge carriers present. The bonding in a molecule is covalent and involves the sharing of atoms and there is no charge separation.

Solutions Chemistry 120 Electrolytes A solute that, when dissolved, produces a solution that conducts is termed an electrolyte, which may be strong or weak. A strong electrolyte is one which is fully dissociated in solution into ions A weak electrolyte is one which is only partially dissociated.

15 Solutions Chemistry 120 Moles and solutions When a substance is dissolved in a solvent, we relate the quantity of material dissolved to the volume of the solution through the concentration of the solution. The concentration is simply the number of moles of the material per unit volume: C = n V n = number of moles; V = volume of solvent

Solutions Chemistry 120 Moles and solutions The units of concentration are: C = n = moles V L3 and we define a molar solution as one which has 1 mole per liter. Alternatively, Concentration = Molarity = number of moles volume of solution

Solutions Chemistry 120 Example

4 g of Na2SO4(s) is dissolved in 500 ml of water. What is the concentration of the solution?

16 Solutions Chemistry 120 Example

4 g of Na2SO4(s) is dissolved in 500 ml of water. What is the concentration of the solution?

Formula mass of Na2SO4(s): Molar Atomic Mass of Na: 22.9898 gmol-1

Molar Atomic Mass of S: 32.064 gmol-1

Molar Atomic Mass of O: 15.9994 gmol-1

Solutions Chemistry 120 Example

4 g of Na2SO4(s) is dissolved in 500 ml of water. What is the concentration of the solution?

Formula mass of Na2SO4(s): (2 x 22.9898)+ 32.064+(4x15.9994)=142.041gmol-1

1 mole of Na2SO4(s) = 142.041g

1 /142.041 mole of Na2SO4(s) = 1 g

Solutions Chemistry 120 Example

4 g of Na2SO4(s) is dissolved in 500 ml of water. What is the concentration of the solution?

1 /142.041 mole of Na2SO4(s) = 1 g

4 Therefore 4 g of Na2SO4(s) = /142.041 mole = 2.82 x 10-2 mole

17 Solutions Chemistry 120 Example

4 g of Na2SO4(s) is dissolved in 500 ml of water. What is the concentration of the solution?

2.82 x 10-2 mole is therefore dissolved in 500 ml of water;

So in 1 L, there are 2 x 2.82 x 10-2 mole

Molarity of solution = 5.64 x 10-2 molL-1

Solutions Chemistry 120 Example

The equation for the dissolution of Na2SO4(s) is

+ 2- Na2SO4(s) 2Na (aq) + SO4 (aq) H2O -2 -1 So if we have 5.64 x 10 molL Na2SO4(s), we must -1 + have 1.13 x 10 moles Na (aq)

-2 2- and 5.64 x 10 mol SO4 (aq) as there are 2 Na cations for every sulfate ion

Solutions Chemistry 120 If we change the volume of the solution then we change the concentration:

If the Na2SO4 solution is diluted with 500ml of water, the concentration or molarity would be halved: 2.82 x 10-2 mole is therefore dissolved in 1000 ml of water Molarity = 2.82 x 10-2 molL-1

18 Solutions Chemistry 120 Dissolution on an atomic level. Solids are held together by very strong forces.

o NaCl(s) melts at 801 C and boils at 1465 oC but it dissolves in water at room temperature.

Solutions Chemistry 120 Dissolution on an atomic level.

When we dissolve NaCl(s) in water we break the bonds between ions but make bonds between the ions and the water

Solutions Chemistry 120 Dissolution on an atomic level.

When we dissolve NaCl(s) in water we break the bonds between ions but make bonds between the ions and the water The ions are hydrated or solvated in solution and these bonds between solvent and solute make the dissolution energetically possible If something does not dissolve then the energetics are wrong for it do do so.

19 Solutions Chemistry 120 Solubility rules All ammonium and Group I salts are soluble. All Halides are soluble except those of silver, lead and mercury (I) All Sulfates are soluble except those of barium and lead. All nitrates are soluble. Everything else is insoluble

Solutions Chemistry 120 • Solutions are homogenous mixtures in which the majority component is the solvent and the minority component is the solute • Solutions are normally liquid but solutions of gases in solids and solids in solids are known. • Ionic compounds dissolve in water to give conducting solutions – they are electrolytes

Solutions Chemistry 120 • Electrolytes are either strong or weak depending on the degree of dissociation in solution • Molecular solutions do not conduct as molecules do not dissociate in solution • The concentration or molarity of a solution is defined by C = n = moles V L3 and the units are molL-1 or moldm-3

20 Solutions Chemistry 120 • When ionic substances dissolve, bonds between particles in the solid break and bonds between the solvent and the ions are made • There are general rules for the solubilities of ionic compounds

Reactions in Solution Chemistry 120 Reactions in solution include • Acid – base reactions • Precipitation reactions • Oxidation- reduction reactions

Reactions in Solution Chemistry 120 Reactions and equilibria

Reactions are often written as proceeding in one direction only – with an arrow to show the direction of the chemical change, reactants to products.

Not all reactions behave in this manner and not all reactions proceed to completion.

Even those that do are dynamic.

21 Reactions in Solution: Acid - Base Chemistry 120 A saturated solution of NaI is placed in contact 131 with Na I(s), which is radioactive.

+ - + Na I I- Na I- - + I Na Na+ - - I- I I + + + - Na Na NaI I Na (aq) - + + I Na Na Na+ - - I I Na+ Na+ + + Na Na I- I- - I + - - I I + Na - Na I Na+ * NaI (s)

Reactions in Solution: Acid - Base Chemistry 120 A saturated solution of NaI is placed in contact 131 with Na I(s), which is radioactive.

+ - + Na I I- Na I- After time, the activity - + I Na Na+ - - I- I I in the solution is + + + - Na Na NaI I Na (aq) - + + I + measured and ...... Na Na Na - - I I Na+ Na+ + + Na Na I- I- - I + - - I I + Na - Na I Na+ * NaI (s)

Reactions in Solution: Acid - Base Chemistry 120 Radioactivity is found in the solution, even though - the concentration of I (aq) has not changed.

+ - + + - + Na I I- Na I- Na I I- Na I- - + - + I Na Na+ I Na Na+ - - - - I- I I I- I I + + + + + + - Na Na - Na Na I Na I Na - - + + I + + I Na Na Na+ Na Na Na+ - - - I - I I Na+ Na+ I Na+ Na+ + + + Na + Na Na I- I- Na I- I- - - I + - I + - - I - I I + Na I + Na - Na - Na I Na+ I Na+

22 Reactions in Solution: Acid - Base Chemistry 120 The equilibrium here is composed of two reactions:

131 + 131 - Na I(s) Na (aq) + I (aq) H2O

+ - Na (aq) + I (aq) NaI(s) H2O

So we write + - NaI(s) Na (aq) + I (aq) H2O

Reactions in Solution: Acid - Base Chemistry 120 Such reactions are termed equilibria and all chemical reactions are equilibria. The symbol for an equilibrium is a double-headed arrow Forward reaction

23 Reactions in Solution: Acid - Base Chemistry 120 Such reactions are termed equilibria and all chemical reactions are equilibria. The symbol for an equilibrium is a double-headed arrow Forward reaction

Reverse reaction

Reactions in Solution: Acid - Base Chemistry 120 Equilibria are important in the chemistry of acids and bases Strong acids and bases are completely ionized But..... Weak acids and bases are not.

24 Reactions in Solution: Acid - Base Chemistry 120 The Arrhenius definition of acid and bases are: an acid is a compound which dissolves in water or reacts with water to give hydronium ions, H O+ 3 (aq) Svante Arrhenius (1859 – 1927) a base is a compound which dissolves in water or reacts with water to give - hydroxide ions, OH (aq)

Reactions in Solution: Acid - Base Chemistry 120 A strong acid is a compound which dissolves and dissociates completely in water or reacts with water + to give hydronium ions, H3O (aq) + - HCl(g) H3O (aq) + Cl (aq) H2O - the double arrow implies that the reaction can go both ways – it is an equilibrium. As a strong acid, the reaction is completely on the RHS: + - HCl(g) H3O (aq) + Cl (aq) H2O

Reactions in Solution: Acid - Base Chemistry 120 A strong base is a compound which dissolves and dissociates completely in water or reacts with water - to give hydroxide ions, OH (aq)

+ - NaOH(s) Na (aq) + OH (aq) H2O Again, we could write this reaction as an equilibrium with a double headed arrow, but the base is strong and the reaction is completely over to the right hand side.

25 Reactions in Solution: Acid - Base Chemistry 120 In a reaction such as + - MeCO2H H3O (aq) + MeCO2 (aq) H2O we write the reaction as going from LHS to RHS. Chemical reactions run both ways, so in this reaction, there are two reactions present: Ionization + - MeCO2H H3O (aq) + MeCO2 (aq) H2O Recombination + - H3O (aq) + MeCO2 (aq) MeCO2H H2O

Reactions in Solution: Acid - Base Chemistry 120

We write the reaction for acetic acid, MeCO2H, as an equilibrium to include the ionization and recombination. + - Ionization MeCO2H H3O (aq) + MeCO2 (aq) H2O

+ - Recombination H3O (aq) + MeCO2 (aq) MeCO2H H2O

+ - MeCO2H H3O (aq) + MeCO2 (aq) H2O As the amount of ionization and recombination are the same, the concentrations of the ions and the molecular form are constant

Reactions in Solution: Acid - Base Chemistry 120 In solution, weak acids establish an equilibrium between the un-ionized or molecular form and the ionized form:

+ - MeCO2H H3O (aq) + MeCO2 (aq) H2O un-ionized ionized molecular form

26 Reactions in Solution: Acid - Base Chemistry 120 In solution, strong acids are completely ionized and even though there is an equilibrium, it lies entirely on the RHS and recombination is negligible:

+ - HBr(g) H3O (aq) + Br (aq) H2O un-ionized ionized molecular form

Reactions in Solution: Acid - Base Chemistry 120 Acids with more than one ionizable hydrogen are termed Polyprotic The common polyprotic acids are

H3PO4 Phosphoric acid

H2SO4 Sulfuric acid

Reactions in Solution: Acid - Base Chemistry 120 Polyprotic acids can ionize more than once

+ - H PO H2SO4(aq) H3O (aq) + HSO4 (aq) 3 4 H2O

- + 2- HSO4 (aq) H3O (aq) + SO4 (aq) H2O

2- + 2- HPO4 (aq) H3O (aq) + PO4 (aq) H2O

Each proton is ionizable and the anions, - dihydrogen phosphate (H2PO4 (aq)) 2- and hydrogen phosphate (HPO4 (aq)) both act as acids, though H3PO4 is a weak acid.

27 Reactions in Solution: Acid - Base Chemistry 120 Polyprotic acids can ionize more than once

+ - H PO H2SO4(aq) H3O (aq) + HSO4 (aq) 3 4 H2O

- + 2- HSO4 (aq) H3O (aq) + PO4 (aq) H2O

2- + 2- HPO4 (aq) H3O (aq) + PO4 (aq) H2O

+ - H SO H2SO4(aq) H3O (aq) + HSO4 (aq) 2 4 H2O

- + 2- HSO4 (aq) H3O (aq) + PO4 (aq) H2O

Reactions in Solution: Acid - Base Chemistry 120

In contrast, H2SO4 is a strong acid and hydrogen - sulfate (HSO4 (aq)) is also a strong acid.

+ - H2SO4(aq) H3O (aq) + HSO4 (aq) H2O

- + 2- HSO4 (aq) H3O (aq) + PO4 (aq) H2O

Reactions in Solution: Acid - Base Chemistry 120 Strong or weak? All acids can be assumed to be weak except the following:

HCl(aq) hydrochloric acid

HBr(aq) hydrobromic acid

HI(aq) hydriodic acid

HClO4(aq) perchloric acid

HNO3(aq) nitric acid

H2SO4(aq) sulfuric acid

28 Reactions in Solution: Acid - Base Chemistry 120 Hydrogens attached to carbon are not ionizable in water

Acetic acid, MeCO2H (or CH3CO2H) has the H structure O

H O H H

Reactions in Solution: Acid - Base Chemistry 120 Only the hydrogen attached to oxygen is ionized in aqueous solution

H H O O O H H + H H O H O H 2 O H H H

The methyl hydrogens are NOT ionizable in aqueous solution.

Reactions in Solution: Acid - Base Chemistry 120 Strong bases are those which ionize in solution of react to generate hydroxide ion. The common strong bases are those which already contain the OH- ion in the solid. 2 Li 3 3 Na11 Mg1 Strong bases 2 4 K Ca are therefore 19 20 5 Rb Sr the hydroxides 37 38 6 Cs Ba of the group I 55 56 and II metals

29 Reactions in Solution: Acid - Base Chemistry 120 Weak bases are the majority and are usually amines and ammonia. These react with water and deprotonate it, forming hydroxide ion and an ammonium ion: H

N N + OH- CH3 CH3 H3C H3C CH3 H2O CH3

Trimethylamine Trimethylammonium

Reactions in Solution: Acid - Base Chemistry 120 Neutralization reactions and titrations Hydroxide and hydronium ions will react to form water. + - H3O (aq) + OH (aq) 2H2O(l) From the stoichiometry of the balanced equation, the hydroxide and hydronium react in a 1:1 ratio.

We can therefore neutralize a known concentration of base or acid with the same quantity of acid or base. This is an example of a titration.

Reactions in Solution: Acid - Base Chemistry 120 Neutralization reactions and titrations We use an indicator to determine the acidity or basicity of a solution:

An indicator is a compound which changes color strongly at a certain level of acidity.

30 Reactions in Solution: Acid - Base Chemistry 120 Neutralization reactions and titrations We add acid or base – the titrant - to a solution of unknown concentration containing a few drops of the indicator solution. When the solution is still acid, no color change occurs; when the indicator changes color, we know the equivalence point – the point where the acidity or basicity has been neutralized. By knowing the concentration and the volume of the titrant, we can calculate the concentration of the of the unknown solution.

Reactions in Solution: Acid - Base Chemistry 120 Decompostion in acid

A solid base, such as Ca(OH)2(s), will dissolve with reaction in an acid. The anion, hydroxide, reacts with the acid to form the calcium salt of the acid:

Ca(OH)2(s) + H2SO4(aq) Ca2SO4(s) + H2O(l)

Reactions in Solution: Acid - Base Chemistry 120 Decompostion in acid

A solid base, such as Ca(OH)2(s), will dissolve with reaction in an acid. The anion, hydroxide, reacts with the acid to form the calcium salt of the acid:

Ca(OH)2(s) + H2SO4(aq) Ca2SO4(s) + H2O(l) Is this balanced?

31 Reactions in Solution: Acid - Base Chemistry 120 Decompostion in acid

A solid base, such as Ca(OH)2(s), will dissolve with reaction in an acid. The anion, hydroxide, reacts with the acid to form the calcium salt of the acid:

Ca(OH)2(s) + H2SO4(aq) Ca2SO4(s) + H2O(l) Is this balanced? No

Reactions in Solution: Acid - Base Chemistry 120 Decompostion in acid

A solid base, such as Ca(OH)2(s), will dissolve with reaction in an acid. The anion, hydroxide, reacts with the acid to form the calcium salt of the acid:

Ca(OH)2(s) + H2SO4(aq) Ca2SO4(s) + H2O(l) Is this balanced? No

2Ca(OH)2(s) + H2SO4(aq) Ca2SO4(s) + 2H2O(l)

Reactions in Solution: Acid - Base Chemistry 120 Decompostion in acid Some anions also decompose in acid. These are usually anions which are derived from gases which are not soluble in water: 2- CO3 (aq) carbonate CO2(g) - HCO3 (aq) hydrogen carbonate CO2(g) 2- S (aq) sulfide H2S(g) - HS (aq) hydrogen sulfide H2S(g) 2- SO3 (aq) sulfite SO2(g) - HSO3 (aq) hydrogen sulfite SO2(g)

32 Gases Chemistry 120 Properties of Gases Kinetic Molecular Theory of Gases Pressure Boyle’s and Charles’ Law The Ideal Gas Law Gas reactions Partial pressures

Gases Chemistry 120 Properties of Gases All elements will form a gas at some temperature Most small molecular compounds and elements are either gases or have a significant vapor pressure.

1 H He 1 Room Temperature Gases 2 2 Li Be B C N O F Ne 3 4 5 6 7 8 9 10 3 Na Mg Al Si P S Cl Ar 11 12 13 14 15 16 17 18 4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 55 56 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

Gases Chemistry 120 Properties of Gases As the temperature rises, all elements form a gas at some point. In the following diagram, Blue represents solids Greenrepresents liquids Red represents gases At O K, all elements are solids At 6000 K, all are gases

33 Gases Chemistry 120

Gases Chemistry 120 Properties of Gases Gases have no shape and no volume. They take the volume and shape of the container Their densities are low – usually measured in gL-1 The atoms or molecules of the gas are far further apart than in a solid or a liquid.

Gases Chemistry 120 Gases as an ensemble of particles The attractive forces between liquids and solids are very strong LiF: M.p.: 848°C B. p.: 1676°C

Solid Liquid Liquid Gas In a gas, the forces between particles are negligible and as there are no attractive forces, a gas will occupy the volume of the container.

34 Gases Chemistry 120 Gases as an ensemble of particles The structures of liquids and solids are well ordered on a microscopic level

Ethanol, C H OH CaCl2 2 5

Gases Chemistry 120 Gases as an ensemble of particles In a gas, there is no order and all the properties of the gas are isotropic – all the properties of the gas are the same in all directions. Gas particles are distributed uniformly throughout the container. They can move throughout the container in straight line trajectories.

Gases Chemistry 120 Gases as an ensemble of particles The directions of the motions of the gas particles are random and The velocities form a distribution – there is a range of possible velocities around an average value. The trajectories of the gas particles are straight lines and there are two possible fates for a gas molecule......

35 Gases Chemistry 120 Gases as an ensemble of particles A gas particle can collide with – the walls of the container Or – another gas molecule When this happens, the gas particle changes direction.

Gases Chemistry 120 Gases as an ensemble of particles Kinetic energy can be transferred between the two colliding particles – one can slow down and the other speed up – but the net change in kinetic energy is zero. These collisions are termed elastic, meaning that there is no overall change in kinetic energy.

Gases Chemistry 120 The average kinetic energy for a given gas is determined by the temperature alone and the width and peak maximum is also determined by the temperature.

The Maxwell-Boltzmann distribution for He

36 Gases Chemistry 120 Gases as an ensemble of particles The force exerted by the gas particles on the walls of the container gives rise to the pressure of the gas. We define pressure as the force exerted per unit area: P = Force = F Area A The unit of pressure is the Pascal (Pa) 1 Pa = 1 Nm-2 In practice, the Pascal is too small - kPa or GPa

Gases Chemistry 120 Pressure measurement Pressure is also measured in several other non – SI units: In industry: Pounds per square in (p.s.i.) In research: Pascal, atmosphere, bar, Torr

Gases Chemistry 120 Pressure conversion factors Atmospheric pressure = 101,325 Pa 1 Atmosphere = 101,325 Pa = 1 bar 1 Atmosphere = 101,325 Pa = 1 bar = 760 Torr = 760 mmHg = 14.7 p.s.i.

37 Gases Chemistry 120 Pressure Measurement Pressure is measured using a manometer or barometer – either one containing Hg or an electronic gauge A mercury manometer is a U–tube connected to the gas vessel, with the other end either evacuated or open to the atmosphere. The measurement of the height difference between the mercury levels on both sides of the ‘U’ gives the pressure......

Gases Chemistry 120 Pressure Measurement Let the height difference between the two Hg levels be ∆h Then the gas pressure is given by

Pgas = P0 + ∆h As P = Force = F = mg where g = 9.81 ms-2 Area AA

Gases Chemistry 120 Pressure Measurement How is the height difference related to the pressure? As density, ρ = m V Then m = ρ V

The volume of the column of mercury is

V = A.∆h

And so m = ρ V = ρ A.∆h

38 Gases Chemistry 120 Pressure Measurement

The pressure above the baseline pressure P0 is therefore

Pgas = mg =ρgA.∆h = ρg∆h AA

The Gas Laws Chemistry 120

The factors that control the behavior of a gas are • The nature of the gas

39 The Gas Laws Chemistry 120

The factors that control the behavior of a gas are • The nature of the gas • The quantity of the gas - n

The Gas Laws Chemistry 120

The factors that control the behavior of a gas are • The nature of the gas • The quantity of the gas - n • The pressure - P

The Gas Laws Chemistry 120

The factors that control the behavior of a gas are • The nature of the gas • The quantity of the gas - n • The pressure - P • The temperature - T

40 The Gas Laws Chemistry 120

The factors that control the behavior of a gas are • The nature of the gas • The quantity of the gas - n • The pressure - P • The temperature - T • The volume of a gas -V

The Gas Laws Chemistry 120

These laws apply to a perfect gas or and ideal gas. All gases behave as ideal gases at ordinary temperatures and pressures. The qualities of an ideal gas are: • Zero size to the gas particles We assume that the volume of the container is very much larger than the total volume of the gas molecules • No attractive forces between atoms

The Gas Laws Chemistry 120

These laws apply to a perfect gas or and ideal gas. All gases behave as ideal gases at ordinary temperatures and pressures. At low temperatures and high pressures gases deviate from ideality. The ideal gas laws are based on three interdependent laws – Boyle’s Law, Charles’ Law and Avogadro’s Law.

41 The Gas Laws Chemistry 120

Boyle’s Law Robert Boyle experimented with gases in Oxford in 1660. He discovered that the product of the volume and the pressure of a gas is a constant, so long as the quantity of gas and the temperature are constant.

The Gas Laws Chemistry 120

Boyle’s Law Mathematically, PV = a constant

as long as n and T are constant

The Gas Laws Chemistry 120

Boyle’s Law Mathematically, PV = a constant, k or P = k V as long as n and T are constant.

42 The Gas Laws Chemistry 120

Boyle’s Law A graph of Boyle’s data shows this relationship:

PV = k

The Gas Laws Chemistry 120

Boyle’s Law

1 A graph of /P as the abscissa and V as the ordinate. V = k P

The graph shows a straight line of slope k

The Gas Laws Chemistry 120

Boyle’s Law

1 As the pressure rises, /P becomes smaller and the graph passes through the origin. This implies that at infinitely large pressure, the volume of a gas is zero. We know that molecules and and atoms have a definite volume, so Boyle’s law must fail at very high pressures.

43 The Gas Laws Chemistry 120

Charles’ Law Jacques Charles was a Feench scientist and aeronaut who discovered (1787) that all gases expand by the same amount when the temperature of the gas rises by the same amount.

The Gas Laws Chemistry 120

Charles’ Law Mathematically, we express this as V = k’T And a graph of Charles’ Law is a straight line:

The Gas Laws Chemistry 120 The Combined Gas Law for a Perfect Gas Combining Boyle’s Law, Charles’ Law and Avogadro’s Law, V = k and V = k’T and V = k”n P

we can say that V ∝ nT P

44 The Gas Laws Chemistry 120 The Combined Gas Law for a Perfect Gas V ∝ nT P Or V =K nT P Rearranging we find

PV = a constant nT

The Gas Laws Chemistry 120 The Combined Gas Law for a Perfect Gas The constant is termed the Universal Gas Constant, R, and takes the value R = 8.314 Jmol-1K-1 So the Universal Gas Law is written as PV = nRT This Law applies to all gases as long as they fulfill the conditions for near ideal behavior – not at high pressure and not at low temperature

The Gas Laws Chemistry 120 Using the Combined Gas Law If the quantity of gas is the same, then changes in pressure, temperature or volume can be calculated easily as

P1V1 = n = P2V2 RT1 RT2

P1V1 = P2V2 T1 T2

45 The Gas Laws Chemistry 120 Using the Combined Gas Law The advantage of this expression is that the units do not matter; the units used for P1 ,V1, and T1 will be returned in the calculation for P2 ,V2, and T2. However, if you have to use PV = nRT, you must use the correct units which are consistent with R. The easiest way is to convert all temperatures to K, all pressures to Pa and all volumes to m3; the value for R is then 8.314 Jmol-1K-1

The Gas Laws Chemistry 120 The absolute temperature scale From Charles’ Law, the decrease in volume per unit temperature is always the same and therefore there must be a minimum temperature that can be reached. This is absolute zero O K, and is the zero point for the absolute temperature scale. The temperature in K is related to the temperature in oC through T/K = T/oC + 273.16

The Gas Laws Chemistry 120 Example: Molecular Mass determinations If we know the mass of gas in a sample of known volume, pressure and temperature, then we can calculate the relative molecular mass as we can calculate n. As n = m then, PV = mRT , so RMM = mRT RMM RMM PV RMM = mRT PV

46 The Gas Laws Chemistry 120 Example: Molar volumes From Avogadro’s Law, equal quantities of gas occupy equal volumes. The volume of one mole of gas is therefore independent of the nature of the gas, as long as the gas behaves as ideal. One mole of a perfect gas at 0oC and 1 atm pressure occupies 22.4 L

The Gas Laws Chemistry 120 Example: Volumes and moles When we react solids or liquids, the easiest way is to measure the mass of the sample and then convert to moles by dividing by the relative molecular mass. For gases, the easiest way is to measure the pressure or the volume, as the densities of gases are so low. For these calculations, you must use the same temperatures and pressures for each gas.

The Gas Laws Chemistry 120 Partial pressures In a mixture of gases, we can measure the total pressure of the mixture – PTotal and therefore we can use PV = nRT to determine the total number of moles of gas present. As the mixture contains more than one gas, we can write the contribution of the pressure of each gas to the total pressure

47 The Gas Laws Chemistry 120 Partial pressures

So the total pressure Ptotal is written as the sum of all the individual pressures of the components of the gas mixture:

The Gas Laws Chemistry 120 Partial pressures

So the total pressure Ptotal is written as the sum of all the individual pressures of the components of the gas mixture:

PTotal = P1 + P2 + P3 + P4 + ......

The Gas Laws Chemistry 120 Partial pressures

So the total pressure Ptotal is written as the sum of all the individual pressures of the components of the gas mixture:

PTotal = P1 + P2 + P3 + P4 + ...... As PV = nRT then

48 The Gas Laws Chemistry 120 Partial pressures

So the total pressure Ptotal is written as the sum of all the individual pressures of the components of the gas mixture:

PTotal = P1 + P2 + P3 + P4 + ...... As PV = nRT then nTotalRT = n1RT + n2RT + n3RT + n4RT + ......

The Gas Laws Chemistry 120 Partial pressures So

PTotal = P1 + P2 + P3 + P4 + ...... nTotalRT = n1RT + n2RT + n3RT + n4RT + ......

The Gas Laws Chemistry 120 Partial pressures So

PTotal = P1 + P2 + P3 + P4 + ...... nTotalRT = n1RT + n2RT + n3RT + n4RT + ......

nTotal = n1 + n2 + n3 + n4 + ......

49 The Gas Laws Chemistry 120 Partial pressures So the pressures of each component of the gas mixture correlate with the number of moles of the gas component of the mixture – a simple extension of Avogadro’s Law.

The Gas Laws Chemistry 120 Partial pressures We can also write the fraction of the total pressure that is due to one of the component:

PTotal = P1 + P2 + P3 + P4 + ......

The Gas Laws Chemistry 120 Partial pressures We can also write the fraction of the total pressure that is due to one of the component:

PTotal = P1 + P2 + P3 + P4 + ......

nTotal = n1 + n2 + n3 + n4 + ......

50 The Gas Laws Chemistry 120 Partial pressures We can also write the fraction of the total pressure that is due to one of the component:

PTotal = P1 + P2 + P3 + P4 + ......

nTotal = n1 + n2 + n3 + n4 + ......

P1 = n1RT

The Gas Laws Chemistry 120 Partial pressures We can also write the fraction of the total pressure that is due to one of the component:

PTotal = P1 + P2 + P3 + P4 + ......

nTotal = n1 + n2 + n3 + n4 + ......

P1 = n1RT

So, P1 = n1

PTotal n1 + n2 + n3 + n4 + ......

The Gas Laws Chemistry 120 Partial pressures

P1 = n1

PTotal n1 + n2 + n3 + n4 + ...... The fraction on the RHS is called the mole fraction and is written as x1 so we can write

P1 = n1 PTotal

n1 + n2 + n3 + n4 + ...... Or P1 = x1 PTotal

51 Thermochemistry Chemistry 120 Energy Energy is defined as the ability to do work. There are several forms of energy Kinetic energy – energy due to motion

Thermochemistry Chemistry 120 Energy Energy is defined as the ability to do work. There are several forms of energy

1 2 Kinetic energy – energy due to motion EK = /2mv Potential energy – the energy due to the position of a particle in a field e.g. Gravitational, electrical, magnetic etc.

Thermochemistry Chemistry 120 Energy The unit of energy is the Joule (J) and 1 J = 1 kgm2s-2 Thermochemistry is the study of chemical energy and of the conversion of chemical energy into other forms of energy. It is part of thermodynamics – the study of the flow of heat.

52 Thermochemistry Chemistry 120 Thermochemically, we define the system as the part of the universe under study and the surroundings as everything else. Systems come in three forms: Open The system can exchange matter and energy with the surroundings Closed The system can exchange energy only with the surroundings Isolated There is no exchange of matter or of energy with the surroundings

Thermochemistry Chemistry 120 Matter is continually in motion and has an internal energy that is composed of several different types There is Translation Rotation Vibration Potential between molecules and inside molecules. The internal energy is written as U

53 Thermochemistry Chemistry 120 Heat is the transfer of energy between the surroundings and the system or between systems. The direction of the heat flow is indicated by the temperature – heat flows along a Temperature gradient from high temperature to low temperature. When the temperature of the system and that of the surroundings are equal, the system is said to be in thermal equilibrium

Thermochemistry Chemistry 120 Heat is the transfer of energy between the surroundings and the system or between systems. The direction of the heat flow is indicated by the temperature – heat flows along a Temperature gradient from high temperature to low temperature. When the temperature of the system and that of the surroundings are equal, the system is said to be in thermal equilibrium